你可以试试:

In [1]: s = pd.DataFrame('a'=[1,2,5, np.nan, np.nan,3],'b'=[1,3, np.nan, np.nan,3,np.nan])

In [4]: s.isna().sum()   
Out[4]: out = {'a'=2, 'b'=3} # the number of NaN values for each column

如果需要nan的总和:

In [5]: s.isna().sum().sum()
Out[6]: out = 5  #the inline sum of Out[4] 

希望这能有所帮助，

import pandas as pd
import numpy as np
df = pd.DataFrame({'a':[1,2,np.nan], 'b':[np.nan,1,np.nan],'c':[np.nan,2,np.nan], 'd':[np.nan,np.nan,np.nan]})

df.isnull().sum()/len(df) * 100

Thres = 40
(df.isnull().sum()/len(df) * 100 ) < Thres

如果只是在pandas列中计算nan值，这里是一个快速的方法

import pandas as pd
## df1 as an example data frame 
## col1 name of column for which you want to calculate the nan values
sum(pd.isnull(df1['col1']))

我写了一个简短的函数(Python 3)来生成.info作为pandas数据框架，然后可以写入excel:

df1 = pd.DataFrame({'a':[1,2,np.nan], 'b':[np.nan,1,np.nan]}) 
def info_as_df (df):
    null_counts = df.isna().sum()
    info_df = pd.DataFrame(list(zip(null_counts.index,null_counts.values))\
                                         , columns = ['Column', 'Nulls_Count'])
    data_types = df.dtypes
    info_df['Dtype'] = data_types.values
    return info_df
print(df1.info())
print(info_as_df(df1))

这使:

<class 'pandas.core.frame.DataFrame'>
RangeIndex: 3 entries, 0 to 2
Data columns (total 2 columns):
 #   Column  Non-Null Count  Dtype  
---  ------  --------------  -----  
 0   a       2 non-null      float64
 1   b       1 non-null      float64
dtypes: float64(2)
memory usage: 176.0 bytes
None
  Column  Nulls_Count    Dtype
0      a            1  float64
1      b            2  float64

使用isna()方法(或者它的别名isnull()，这也兼容较旧的pandas版本< 0.21.0)，然后求和来计算NaN值。其中一列:

>>> s = pd.Series([1,2,3, np.nan, np.nan])

>>> s.isna().sum()   # or s.isnull().sum() for older pandas versions
2

对于一些列，这也适用:

>>> df = pd.DataFrame({'a':[1,2,np.nan], 'b':[np.nan,1,np.nan]})

>>> df.isna().sum()
a    1
b    2
dtype: int64

import pandas as pd
import numpy as np

# example DataFrame
df = pd.DataFrame({'a':[1,2,np.nan], 'b':[np.nan,1,np.nan]})

# count the NaNs in a column
num_nan_a = df.loc[ (pd.isna(df['a'])) , 'a' ].shape[0]
num_nan_b = df.loc[ (pd.isna(df['b'])) , 'b' ].shape[0]

# summarize the num_nan_b
print(df)
print(' ')
print(f"There are {num_nan_a} NaNs in column a")
print(f"There are {num_nan_b} NaNs in column b")

给出输出:

     a    b
0  1.0  NaN
1  2.0  1.0
2  NaN  NaN

There are 1 NaNs in column a
There are 2 NaNs in column b